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Chapter 1: Problem 6

Given \(f(t)=\frac{|3+t|-|t|-3}{t}\) express \(f(t)\) without absolute-value bars if (a) \(t>0\); (b) \(-3 \leq t<0 ;\) (c) \(t<-3\)

Short Answer

Expert verified
For \( t > 0, \) \( f(t) = 0 \). For \( -3 \leq t < 0, \) \( f(t) = 2 \). For \( t < -3, \) \( f(t) = -\frac{6}{t} \).

Step by step solution

01

- Analyze the function for different intervals of t

The function given is \( f(t) = \frac{|3+t| - |t| - 3}{t} \). Since it contains absolute values, the function behavior changes based on the value of \( t \).
02

- Case (a): When \( t > 0 \)

For \( t > 0 \), \( |t| = t \). Then, analyze \( |3 + t| \). Since \( t > 0 \) and \( 3 + t > 3 \), \( |3 + t| = 3 + t \). Substituting in, we get \[ f(t) = \frac{3 + t - t - 3}{t} = \frac{0}{t} = 0 \]
03

- Case (b): When \( -3 \leq t < 0 \)

For \( -3 \leq t < 0 \), \( |t| = -t \). Analyze \( |3 + t| \). Since \( -3 \leq t < 0 \) implies \( 0 \leq 3 + t < 3 \), \( |3 + t| = 3 + t \). Substituting in, we get \[ f(t) = \frac{3 + t - (-t) - 3}{t} = \frac{3 + t + t - 3}{t} = \frac{2t}{t} = 2 \]
04

- Case (c): When \( t < -3 \)

For \( t < -3 \), \( |t| = -t \). Analyze \( |3 + t| \). Since \( t < -3 \) implies \( 3 + t < 0 \), \( |3 + t| = -(3 + t) = -3 - t \). Substituting in, we get \[ f(t) = \frac{-3 - t - (-t) - 3}{t} = \frac{-3 - t + t - 3}{t} = \frac{-6}{t} = -\frac{6}{t} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Absolute Value Functions
Absolute value functions are essential in many areas of mathematics, including calculus. The absolute value of a number is its distance from zero on the number line, regardless of direction. It is always non-negative. For any real number, the absolute value is denoted by vertical bars, like this: \( |x| \).
The absolute value function can be defined piecewise as:
  • \( |x| = x \text{, if } x \text{ is greater than or equal to } 0 \)
  • \( |x| = -x \text{, if } x \text{ is less than } 0 \)
In the given problem, we encounter absolute values in the expression \( f(t) = \frac{|3+t| - |t| - 3}{t} \). As shown in the steps, we need to consider different intervals for \( t \) to simplify the absolute values.
Piecewise Functions
Piecewise functions are defined by different expressions depending on the interval of the input variable. They are particularly useful to describe situations where a function exhibits different behaviors over different ranges.
In the case of our exercise, we analyze the function \( f(t) \) in three different intervals:
  • For \( t > 0 \)
  • For \( - 3 \leq t < 0 \)
  • For \( t < -3 \)
By breaking down the problem this way, it becomes easier to manage the absolute values and simplify the function. Each interval creates a different version of the function where the absolute value expressions are easier to handle, allowing us to find a simplified form.
Interval Analysis
Interval analysis is a technique used to examine the behavior of functions over different segments of their domain. By dividing the domain into smaller sub-intervals, we can understand how different parts of a function behave.
Here's how we applied interval analysis to our exercise:
  • **For** \( t > 0 \)
    • By substituting the conditions of this interval into \( |3 + t| \) and \( |t| \), the function simplifies as follows:

    \( f(t) = \frac{3 + t - t - 3}{t} = \frac{0}{t} = 0 \)

  • **For** \( -3 \leq t < 0 \)
    • With the conditions in this interval, the function modification:

    \( f(t) = \frac{3 + t - (-t) - 3}{t} = \frac{3 + t + t - 3}{t} = \frac{2t}{t} = 2 \)

  • **For** \( t < -3 \)
    • Substituting the intervals values, we get:

    \( f(t) = \frac{-3 - t - (-t) - 3}{t} = \frac{-3 - t + t - 3}{t} = \frac{-6}{t} = -\frac{6}{t}\)

    By using interval analysis, we ensure the function's behavior is fully understood and can be expressed without the complications of absolute value.

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Most popular questions from this chapter

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